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Ed
 Post subject: Assassin 423
PostPosted: Mon Aug 15, 2022 6:44 pm 
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Posts: 1044
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a423.png
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Note: 1-9 cannot repeat on either diagonal. Note, one disjoint 16(3) in n2.

Assassin 423

My last A for now. Back to Revisits on the 2nd and 16th unless someone posts an Assassin. This one started as a 1.50 but couldn't solve it. Changed it to a 1.70 (easier for Jsudoku) and solved it! Hard one. Found one nice step so it (only just) meets my criteria of: must be interesting.
triple click code:
3x3:d:k:3840:2817:2817:2818:5379:5379:5379:5379:5380:4869:3840:2817:2818:2818:4120:5380:5380:3847:4869:4869:3840:4120:3607:4120:7432:3847:3847:2569:2569:2569:7432:3607:7432:6666:5899:3847:3852:3852:4365:4365:7432:6666:5899:5899:5899:1294:1294:4365:7432:6666:6666:2831:2831:2831:5392:5392:7432:6666:2833:4882:4115:3860:3860:5392:2069:2069:5654:2833:4882:3590:4115:3860:2069:5654:5654:5654:2833:4882:3590:3590:4115:
solution: 
+-------+-------+-------+
| 1 2 6 | 8 9 3 | 4 5 7 |
| 7 5 3 | 1 2 4 | 6 8 9 |
| 4 8 9 | 7 6 5 | 2 1 3 |
+-------+-------+-------+
| 5 4 1 | 3 8 6 | 9 7 2 |
| 9 6 7 | 2 4 1 | 5 3 8 |
| 2 3 8 | 9 5 7 | 1 6 4 |
+-------+-------+-------+
| 6 7 5 | 4 3 2 | 8 9 1 |
| 8 1 4 | 6 7 9 | 3 2 5 |
| 3 9 2 | 5 1 8 | 7 4 6 |
+-------+-------+-------+
Cheers
Ed
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Andrew
 Post subject: Re: Assassin 423
PostPosted: Fri Aug 19, 2022 4:19 am 
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Thanks Ed for the latest Assassin! Yes, it was quite a hard one.

If you still know what the 1.5 version was, please feel free to post it as a V1.5; I love a challenge, within reasonable limits.

Apart from that, the Revisits will be welcome; the original puzzles were posted so long ago that they can be treated as if they were new puzzles.

Here's how I solved Assassin 423:
Prelims

a) R34C5 = {59/68}
b) R5C12 = {69/78}
c) R6C12 = {14/23}
d) 11(3) cage at R1C2 = {128/137/146/236/245}, no 9
e) 11(3) cage at R1C4 = {128/137/146/236/245}, no 9
f) 21(3) cage at R1C9 = {489/579/678}, no 1,2,3
g) 19(3) cage at R2C1 = {289/379/469/478/568}, no 1
h) 10(3) cage at R4C1 = {127/136/145/235}, no 8,9
i) 11(3) cage at R6C7 = {128/137/146/236/245}, no 9
j) 21(3) cage at R7C1 = {489/579/678}, no 1,2,3
k) 11(3) cage at R7C5 = {128/137/146/236/245}, no 9
l) 19(3) cage at R7C6 = {289/379/469/478/568}, no 1
m) 8(3) cage at R8C2 = {125/134}

1a. 45 rule on N4 1 outie R5C4 = 2 -> R56C3 = 15 = {69/78}
1b. 5 in N4 only in 10(3) cage at R4C1 = {145/235}, 5 locked for R4, clean-up: no 9 in R3C5
1c. Combined cage R6C12 + 11(3) cage = 16(5) = {12346}, locked for R6, 6 locked for N6, clean-up: no 9 in R5C3
1d. 6 in N4 only in R5C123, locked for R5
1e. 5 in N6 only in R5C789, locked for R5
1f. 8(3) cage at R8C2 = {125/134}, 1 locked for N7
1g. 45 rule on N78 2 innies R7C34 = 9 = [27]/{36/45}/[81], no 9, no 7 in R7C3, no 8 in R7C4

2a. 5 in N6 only in 23(4) cage at R4C8 = {1589/3578} (cannot be {2579} = 2{579} which clashes with R5C12), no 2,4, 8 locked for N6
2b. 4 in R5 only in R5C56, locked for N5
2c. 45 rule on N6 2 innies R4C79 = 11 = {29/47}, no 1,3
2d. Hidden killer quad 6,7,8,9 in R5C123 and 23(4) cage, R5C123 contains 6 and two of 7,8,9, 23(4) cage contains 8 and one of 7,9 -> R4C8 = {789}, 7,8,9 locked for R5
2e. 45 rule on D\ 3 innies R4C4 + R5C5 + R6C6 = 14
2f. Min R5C5 + R6C6 = 6 -> max R4C4 = 8
[Outies from D/ = 13 also gives this result.]
2g. R5C5 + R6C6 cannot total 7 -> no 7 in R4C4
2h. R4C4 + R5C5 cannot total 6 -> no 8 in R6C6
2i. 45 rule on N2 3 innies R1C56 + R3C5 = 18 = {189/369/378/459/468/567} (cannot be {279} because R3C5 only contains 5,6,8), no 2 in R1C56
2j. 45 rule on N236 3(2+1/1+2) innies R3C57 + R4C7 = 17
2k. R3C5 + R4C7 cannot total 11,16 -> no 1,6 in R3C7
2l. R3C5 + R4C7 cannot total 14 (which clashes with R34C5, CCC) -> no 3 in R3C7

3a. R34C5 = [59/68/86], R3C57 + R4C7 = 17 (step 2j), R4C4 + R5C5 + R6C6 = 14 (step 2e), R4C79 = {29/47} (step 2c)
3b. R3C57 + R4C7 = [584]/6{29}/6{47}/8{27}/[854]
3c. Consider combinations for R4C4 + R5C5 + R6C6 = 14 = {149/158/167/347/356}
R4C4 + R5C5 + R6C6 = {149}, locked for N5
or R4C4 + R5C5 + R6C6 = {158/167/356} => R5C6 = 4 (hidden single in N5), no 4 in R4C7 => no 5 in R3C5
or R4C4 + R5C5 + R6C6 = {347} = [347] => 10(3) cage at R4C1 = {145}, locked for R4 => R4C79 = {29}, locked for R4
-> R34C5 = {68}, locked for C5
3d. R3C57 + R4C7 = 6{29}/6{47}/8{27}/[854], no 8 in R3C7
3e. R5C5 = {134}, 11(3) cage at R7C5 = {137/245} -> combined half cage R5789C5 = 1{245}/3{245}/4{137}, 4 locked for C5
3f. R1C56 + R3C5 = 18 (step 2i), R3C5 is even, R1C5 odd -> R1C6 must be odd, no 4,6,8 in R1C6
3g. Variable hidden killer pair 2,4 in 11(3) cage at R1C4 and disjoint 16(3) cage at R2C6, 16(3) cage cannot contain both of 2,4 -> 11(3) cage must contain at least one of 2,4 = {128/146/236/245}, no 7
3h. 2 of {236/245} must be in R2C5 -> no 3,5 in R2C5
3i. Killer pair 1,2 in R2C5 and 11(3) cage at R7C5, locked for C5
3j. Killer pair 3,4 in R5C5 and 11(3) cage at R7C5, locked for C5
3k. Min R13C5 = 11 -> max R1C6 = 7
3l. R4C4 + R5C5 + R6C6 = {149/347/356} (cannot be {158/167} because R5C5 only contains 3,4), no 8
3m. 26(5) cage at R4C7 = {14579/24569} (cannot be {13679} because 1,3,6 only in R5C6 + R7C4, cannot be {23579} = [23]{579} when 1 then in R4C46 and R4C7 = 2 clash with 10(3) cage at R4C1), no 3, clean=up: no 6 in R7C3 (step 1g)
3n. R4C4 + R5C5 + R6C6 = {347/356} = [347/635] (cannot be {149} which clashes with R5C6), no 1,9, 3 locked for N5 and D\
3o. 26(5) cage at R4C7 = {14579} (cannot be {24569} = [24956] which clashes with R4C4 + R5C5 + R6C6 = [635], CCC), no 2,6, clean-up: no 9 in R4C9 (step 2c), no 3 in R7C4 (step 1g)

[This 45 was seen earlier but is only helpful after eliminating 1 from R4C4]
4a. 45 rule on R4 4 remaining innies R4C4568 = 24 = {1689/3678}
4b. 1 of {1689} only in R4C6 -> no 9 in R4C6
4c. 9 in R4 only in R4C78, locked for N6
4d. 9 in R5 only in R5C12 = {69}, locked for N4
4e. Naked pair {78} in R56C3, locked for C3, clean-up: no 1 in R7C4 (step 1g)
4f. 26(5) cage at R4C7 (step 3o) = {14579} -> R5C6 = 1, R5C5 = 4 (hidden single in N5), placed for both diagonals, R4C4 = 3, R6C6 = 7 (step 3n), placed for D\, R56C3 = [78], clean-up: no 2 in 10(3) cage at R4C1, no 2 in R7C3 (step 1g)
4g. R5C789 = {358} = 16 -> R4C8 = 7 (cage sum)
4h. Naked triple {145} in 10(3) cage at R4C1, 1,4 locked for N4, 4 locked for R4
4i. Naked pair {23} in R6C12, locked for R6
4j. 11(3) cage at R7C5 = {137}, only remaining combination), locked for N8, 1,7 locked for C5

5a. R4C7 = 9 -> R6C5 = 5, R7C4 = 4, R6C4 = 9, R7C3 = 5, both placed for D/
5b. 45 rule on N8 2 remaining innies R89C4 = 11 = {56}, locked for C4, N8 and 22(4) cage at R8C4)
5c.8(3) cage at R8C2 = {134} (only remaining combination) -> R8C3 = 4, R8C2 + R9C1 = {13}, 3 locked for N7
5d. 21(3) cage at R7C1 = {678} (only remaining combination), 7,8 locked for N7
5e. Naked pair {29} in R9C23, locked for R9
5f. R9C6 = 8 -> R4C6 = 6, placed for D/
5g. R1C9 + R2C8 = [78] -> R2C7= 6 (cage sum)
5h. R34C5 = [68], R1C5 = 9 -> R1C6 = 3 (step 2i)
5i. R1C56 = [93] = 12 -> R1C78 = 9 = {45}, locked for R1 and N3
5j. 11(3) cage at R1C4 = [812]
5k. R2C3 = 3 -> R1C23 = 8 = {26}, locked for N1
5l. R1C1 = 1 -> R2C2 + R3C3 = [59], all placed for D\
5m. 16(3) cage at R7C7 = [826]
5n. 15(3) cage at R7C8 = {159} (only remaining combination) -> R8C9 = 5, R7C89 = {19}, locked for R7, 1 locked for N9

and the rest is naked singles without using the diagonals.
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Ed
 Post subject: Re: Assassin 423
PostPosted: Mon Aug 22, 2022 8:22 pm 
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Location: Sydney, Australia
Attachment:
a423v1.5.png
a423v1.5.png [ 96.32 KiB | Viewed 4898 times ]
Assassin 423 v1.50
Thanks for this Andrew! Haven't finished Andrew's WT yet but this version 1.50 blocks his key step 3c. Same solution.

triple click code:
3x3:d:k:3840:2817:2817:2818:5379:5379:5379:5379:5380:4869:3840:2817:2818:2818:7686:5380:5380:3847:4869:4869:3840:7686:7686:7686:7432:3847:3847:2569:2569:2569:7432:7686:7432:6666:5899:3847:3852:3852:4365:4365:7432:6666:5899:5899:5899:1294:1294:4365:7432:6666:6666:2831:2831:2831:5392:5392:7432:6666:2833:4882:4115:3860:3860:5392:2069:2069:5654:2833:4882:3607:4115:3860:2069:5654:5654:5654:2833:4882:3607:3607:4115:
Cheers
Ed
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wellbeback
 Post subject: Re: Assassin 423
PostPosted: Tue Aug 23, 2022 8:59 pm 
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Thanks Ed. Here's how I did the first one...
Assassin 423 WT:
1. Outies n4 -> r5c4 = 2
-> r56c3 = +15(2)
-> 5 in n4 in 10(3)n4 = {145} or {235}

2. Innies r6 = r6c3456 = +29(4) = {5789} with 5 in n5 (r6c456)
-> 5 in n6 in r5
Also -> 6 in r6 in 11(3)n6 = {146} or {236}
-> 6 in n4 in r5
-> 6 in n5 in r4

3. Innies n6 = r4c79 = +11(2) = {29}, {38}, or {47)
Two of (789) are in n4 in r5
-> Those same two in n6 in r4c789
-> r4c8 from (789)
Also those same two in n5 in r6

Also whichever of (789) is in r6c3 is in n6 in r5
-> is also in n5 in r4

4. Of the pairs (14) and (23} in r4
One of those pairs is in n4/r4
For the other pair in r4, one each must be in n5 and the other in n6
-> Either 10(3)n4 = {145}, r4c79 = {29}, 3 in r4c46
Or 10(3)n4 = {235}, 1 in r4c46, r4c79 = {47}

-> r5c56 from {14} or {34}

5! 11(3)n2 and 11(3)n8 must each contain at least one of (12).
2 already in c4 at r5c4 -> not both 11(3)s can contain a 2 -> at least one of the 11(3)s must contain a 1.
1 in n8 not in 19(3)
-> Either 1 in 11(3)n8 or in r789c4
In the latter case 11(3)n2 must contain a 1 at r2c5
-> 1 locked in c5 in r2789c5

6! Outies D/ = r2c7 + r4c4 + r8c3 = +13(3)
-> 6 cannot be in r4c4 because that puts 6 in D/ in 21(3)n3 which puts min r2c7 = 7 which contradicts Outies D/
-> 6 in r4c56

7! Innies D\ = [r4c4,r5c5,r6c6] = +14(3)
Given previous placements this can only be [149] or [347]
I.e., r5c5 = 4

(This is the point at which the second puzzle needs an alternate path).
In the former case since (59) both in r6 in n5 this puts 14(2)r3c5 = {68}
and the latter case puts H11(2)n6 = {29} which also puts 14(2)r3c5 = {68}

8. -> Only solution for 11(3)n8 = {137}
-> (HS 2 in n8) 19(3)n8 = {289}
-> r789c4 = {456}
Also Innies D\ [r4c4,r5c5,r6c6] = [347]
-> 10(3)n4 = {145} and 5(2)n4 = {23}
-> 11(3)n6 = {146}
Also H11(2)n6 = {29} and 23(4)n6 = {3578} with r4c8 from (78)

9. Since r3c5 from (68) and Innies n2 = +18(3) -> 2 not in Innies n2
-> HS 2 in c5 -> r2c5 = 2
-> r12c4 = {18}
-> r3c4 = 7
Also 14(2)c5 = [68] and r4c6 = 6
etc.

For the second one - starting at Step 7 above...
Assassin 423 (2) WT:
7! Innies D\ = [r4c4,r5c5,r6c6] = +14(3)
Given previous placements this can only be [149] or [347]
I.e., r5c5 = 4

8! [r4c4,r5c5,r6c6] cannot be [149] since that forces both (12) in c6 into n2 leaving no solution for 11(3)n2
-> [r4c4,r5c5,r6c6] = [347]
-> 10(3)n4 = {145} and 5(2)n4 = {23}
-> H11(2)n6 = {29} and 11(3)n6 = {146}
-> r5c6 = 1 and r6c45 = {59}
-> r4c56 = {68}
-> 23(4)n6 = [7{358}]
-> 15(2)n4 = {69} and r45c3 = [78]

9. 7 in r6c6 prevents 3 in 19(3)n8
-> 3 in 11(3)n8
Innies n8 = r789c4 = +15(3) cannot contain both (17)
-> 11(3)n8 must contain at least one of (17)
-> 11(3)n8 = {137}
-> (HS 2 in n8) 19(3)n8 = {289}
-> r4c56 = [86]
Also -> r789c4 = {456}
-> r6c45 = [95]
-> r123c4 = {178}
Since 3 already in c5 -> 11(3)n2 = [{18}2] and r3c4 = 7
Also (IOD n2) r1c56 = +12(2) = [93]
-> r23c6 = {45}

10. Remaining cells 29(6)r3c7 = r3c7,r7c3 = {25}
Innies n78 = r7c34 = +9(2) = [54]
-> r3c7 = 2
-> r4c79 = [92]
etc.
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Andrew
 Post subject: Re: Assassin 423
PostPosted: Fri Sep 16, 2022 7:35 pm 
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At a quick glance, wellbeback's solving path for the original A423 must have been very different from mine, as often happens. V1.5 only required changes to his later steps whereas for me it was almost solving a completely new puzzle.

Here is how I solved Assassin 423 V1.5:
Prelims

a) R5C12 = {69/78}
b) R6C12 = {14/23}
c) 11(3) cage at R1C2 = {128/137/146/236/245}, no 9
d) 11(3) cage at R1C4 = {128/137/146/236/245}, no 9
e) 21(3) cage at R1C9 = {489/579/678}, no 1,2,3
f) 19(3) cage at R2C1 = {289/379/469/478/568}, no 1
g) 10(3) cage at R4C1 = {127/136/145/235}, no 8,9
h) 11(3) cage at R6C7 = {128/137/146/236/245}, no 9
i) 21(3) cage at R7C1 = {489/579/678}, no 1,2,3
j) 11(3) cage at R7C5 = {128/137/146/236/245}, no 9
k) 19(3) cage at R7C6 = {289/379/469/478/568}, no 1
l) 8(3) cage at R8C2 = {125/134}

1a. 45 rule on N4 1 outie R5C4 = 2 -> R56C3 = 15 = {69/78}
1b. 5 in N4 only in 10(3) cage at R4C1 = {145/235}, 5 locked for R4
1c. Combined cage R6C12 + 11(3) cage at R6C7 = 16(5) = {12346}, locked for R6, 6 locked for N6, clean-up: no 9 in R5C3
1d. 6 in N4 only in R5C123, locked for R5
1e. 5 in N6 only in R5C789, locked for R5
1f. 8(3) cage at R8C2 = {125/134}, 1 locked for N7
1g. 45 rule on N78 2 innies R7C34 = 9 = [27]/{36/45}/[81], no 9, no 7 in R7C3, no 8 in R7C4

2a. 5 in N6 only in 23(4) cage at R4C8 = {1589/3578} (cannot be {2579} = 2{579} which clashes with R5C12), no 2,4, 8 locked for N6
2b. 4 in R5 only in R5C56, locked for N5
2c. 45 rule on N6 2 innies R4C79 = 11 = {29/47}, no 1,3
2d. Hidden killer quad 6,7,8,9 in R5C123 and 23(4) cage, R5C123 contains 6 and two of 7,8,9, 23(4) cage contains 8 and one of 7,9 -> R4C8 = {789}, 7,8,9 locked for R5
2e. 45 rule on D\ 3 innies R4C4 + R5C5 + R6C6 = 14
2f. Min R5C5 + R6C6 = 6 -> max R4C4 = 8
[Outies from D/ = 13 also gives this result.]
2g. R5C5 + R6C6 cannot total 7 -> no 7 in R4C4
2h. R4C4 + R5C5 cannot total 6 -> no 8 in R6C6

3a. 45 rule on N236 2 innies R34C7 = 1 outie R4C5 + 3
3b. R3C7 cannot be 3 (IOU)
3c. R34C7 cannot total 4 -> no 1 in R4C5

4a. 2 in N8 only in 11(3) cage at R7C5 = {128/236/245} or 19(3) cage at R7C6 = {289} -> 19(3) cage = {289/379/469/478} (cannot be {568} which clashes with {128/236/245}, blocking cages), no 5
4b. 45 rule on N8 3 innies R789C4 = 15 = {159/168/357/456} (cannot be {348} which clashes with 19(3) cage)
4c. R4C4 + R5C5 + R6C6 = 14 (step 2e) = {149/158/167/347/356} = [149/815/617/347/635]
4d. Consider combinations for R4C4 + R5C5 + R6C6
R4C4 + R5C5 + R6C6 = [149/815/617/635] => R789C4 = some/all of {159/357/456}
or R4C4 + R5C5 + R6C6 = [347] => 19(3) cage = {289/469} => R789C4 = {159/357/456} (cannot be {168} which clashes with 19(3) cage)
-> R789C4 = {159/357/456}, no 8, 5 locked for C4 and N8
4e. 5 in N5 only in R6C56, locked for 26(5) cage at R4C7, no 5 in R7C4, clean-up: no 4 in R7C3 (step 1g)
4f. 5 in C4 only in R89C4, locked for 22(4) cage at R8C4
4g. 1 of {159} must be in R7C4 -> no 1 in R89C4
4h. 45 rule on N7 3 innies R7C3 + R9C23 = 16 = {259/268/349/367} (cannot be {358/457} which clash with 8(3) cage at R8C2
4i. 22(4) cage at R8C4 contains 5 = {2569/2578/3568/4567}
4j. R7C3 + R9C23 = {259/268/367} (cannot be {349} = 3{49} because 22(4) cage cannot contain both of 4,9), no 4
4k. {367} must be 3{67} (cannot be 6{37} because 22(4) cage cannot contain both of 3,7), no 3 in R9C23

5a. R7C34 = 9 (step 1g)
5b. R4C4 + R5C5 + R6C6 (step 4c) = [149/815/617/347/635]
5c. Consider placement for 1 in N8
R7C4 = 1 => R7C3 = 8 => no 8 in R4C4
or 1 in 11(3) cage at R7C5, 1 locked for C5
-> R4C4 + R5C5 + R6C6 cannot be [815]
5d. R4C4 + R5C5 + R6C6 = [149/617/347/635], no 8
5e. R789C4 (step 4d) = {159/357/456}
5f. Consider combinations for R789C4
R789C4 = {159} => R7C4 = 1
or R789C4 = {357/456} => 9 in N8 only in 19(3) cage at R7C6, locked for C6
-> R4C4 + R5C5 + R6C6 cannot be [149]
5g. R4C4 + R5C5 + R6C6 = [617/347/635], no 9, no 1 in R4C4
5h. 45 rule on R4 using R4C79 = 11 (step 2c), 4 remaining innies R4C4568 = 24 must contain 6,8 for R4 = {1689/3678}
5i. 1 of {1689} only in R4C6 -> no 9 in R4C6

6a. R7C34 = [27/54/81] (cannot be {36} which clash with R4C4), no 3,6
6b. R7C4 = {147} -> no 4,7 in R89C4
6c. R7C3 + R9C23 (step 4j) = {259/268}, no 7, 2 locked for N7
6d. 8(3) cage at R8C2 = {134}, 4 locked for N7
6e. Naked triple {134} in R5C5 + R8C8 + R9C9, locked for D/
6f. R4C4568 (step 5h) = {3678}, no 9, 3,7 locked for R4, 3 locked for N5, clean-up: no 2 in 10(3) cage at R4C1
6g. Naked triple {145} in 10(3) cage, 1,4 locked for N4, 4 locked for R4
6h. Naked pair {29} in R4C79, locked for N6, clean-up: no 3 in 11(3) cage at R6C7
6i. 9 in N4 only in R5C12 = {69}, locked for N4
6j. Naked pair {78} in R56C3, locked for C3, clean-up: no 1 in R7C4 (step 1g)
6k. 1 in N8 only in 11(3) cage at R7C5, 1 locked for C5

7a. R5C5 = 4, placed for both diagonals, R5C6 = 1
7b. R4C4 + R5C5 + R6C6 = 14 (step 4c) -> R4C4 + R6C6 = 10 = [37], both placed for D\
7c. R56C3 = [78] -> R6C45 = [95], 9 placed for D/
7d. R5C6 + R6C56 + R7C4 = [1574] = 17 -> R4C7 = 9 (cage sum)
7e. R7C4 = 4 -> R7C3 = 5 (step 1g), placed for D/
7f. 21(3) cage at R1C9 = {678} (only remaining combination), locked for N3
7g. R3C7 + R4C4 + R5C5 + R6C4 + R7C3 = [23495] = 23 -> R4C6 = 6, placed for D/
7h. 19(3) cage at R7C6 = {289} (only remaining combination), locked for C6, 2,8 locked for N8
7i. 21(3) cage at R7C1 = {678} (only remaining combination), 6,8 locked for N7
7j. Naked pair {29} in R9C23, locked for R9 -> R9C6 = 8
7k. 11(3) cage at R7C5 = {137} (only remaining combination), 3,7 locked for C5

8a. R4C5 = 8
8b. 8 in N2 only in 11(3) cage at R1C4 = {128} -> R12C4 = {18}, R2C5 = 2
8c. R3C4 = 7 (hidden single in N2), R4C5 = 8 -> R2C6 + R3C56 = 15 = {456} (only possible combination) -> R3C5 = 6
8d. R1C56 = [93] (hidden pair in N2) = 12 -> R1C56 = 9 = {45}, locked for R1 and N3
8e. R8C3 = 4 (hidden single in N7) -> R4C3 = 1
8f. R3C3 = 9 -> R1C1 + R2C2 = 6 = [15], all placed for D\
8g. R7C7 + R8C8 + R9C9 = [826], R9C3 = 3
8h. 14(3) cage at R8C7 = {347} (only remaining combination) -> R8C7 = 3

and the rest is naked singles, without using the diagonals.
I've now looked at wellbeback's WTs. With hindsight I ought to have spotted his step 5 immediately after my steps 4a and 4b but I must have been focussing on using as many of my original steps as possible. His step 6 is also very powerful; together those two steps pretty much solve these two puzzles.
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